Answer to Question #129508 in Mechanical Engineering for Dipankar Kundu

Question

An ideal Stirling cycle cooler, using helium (considered ideal) as its working fluid, has a compression temperature Tc=32°C and an expansion temperature
has a compression temperature Tc=32°C and an expansion temperature
TE=-21°C.
If state 1, at the beginning of the compression, is defined with a volume V1=40
cm3 and a pressure P1=1.8 MPa, and state 2 is defined with a volume V2=33
cm3, what is the heat qE absorbed during the expansion process (in Joules)?

Answer 1

Given data are

T_H = 32⁰ C + 273 = 305 k

T_L = -21⁰ C + 273 = 252 k

V₁ = 40 cm³ , V₃ = 33 cm³

n = ?

work done in isothermal expansion

w = nRT ln v₂/v₁

= 2.303 *1 *8.31 * 305 log (40/33)

= 487.313 Joules

in isothermal process change in u = 0

from its law of thermodynamics

dq = change in u + change in w

dq = 0 + w

hence,

heat absorbed q = w

q = 487.313 Joules